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CBSE Board Examination 2026 · Section D · 5 marks

Q33: Find A−1A^{-1} and solve a system, or the values of xx that make a matrix singular

Board 2026Ch 4 · DeterminantsCh 3 · Matrices
The question

(a) If

A=[021−2−1−21−10],A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix},

find A−1A^{-1} and use it to solve the following system of equations :

−2y+z=7-2y + z = 7⁠, 2x−y−z=82x - y - z = 8⁠, x−2y=10x - 2y = 10

OR

(b) If

[3−1sin⁡3x−74cos⁡2x−1172]\begin{bmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{bmatrix}

is a singular matrix, then find all values of xx where x∈[0,π2]x \in \left[0, \dfrac{\pi}{2}\right]⁠.

Exam answer

(a)

∣A∣=0(0−2)−2(0+2)+1(2+1)=−4+3=−1≠0\begin{aligned} |A| &= 0(0 - 2) - 2(0 + 2) \\ &\quad + 1(2 + 1) \\ &= -4 + 3 = -1 \ne 0 \end{aligned}
Why?

Expand along the first row: each entry times its 2×22 \times 2 minor, with signs + − ++\,-\,+⁠:

∣A∣=a11M11−a12M12+a13M13\begin{aligned} |A| &= a_{11}M_{11} - a_{12}M_{12} \\ &\quad + a_{13}M_{13} \end{aligned}

For example, deleting row 1 and column 1 leaves

M11=∣−1−2−10∣=0−2M_{11} = \begin{vmatrix} -1 & -2 \\ -1 & 0 \end{vmatrix} = 0 - 2

Written out in full:

∣A∣=0∣−1−2−10∣−2∣−2−210∣+1∣−2−11−1∣\begin{aligned} \lvert A\rvert &= 0\begin{vmatrix} -1 & -2 \\ -1 & 0 \end{vmatrix} - 2\begin{vmatrix} -2 & -2 \\ 1 & 0 \end{vmatrix} \\ &\quad + 1\begin{vmatrix} -2 & -1 \\ 1 & -1 \end{vmatrix} \end{aligned}

Each 2×22 \times 2 determinant is (top left)(bottom right) minus (top right)(bottom left):

∣−1−2−10∣=(−1)(0)−(−2)(−1)=0−2=−2∣−2−210∣=(−2)(0)−(−2)(1)=0+2=2∣−2−11−1∣=(−2)(−1)−(−1)(1)=2+1=3\begin{aligned} \begin{vmatrix} -1 & -2 \\ -1 & 0 \end{vmatrix} &= (-1)(0) - (-2)(-1) \\ &= 0 - 2 = -2 \\[4pt] \begin{vmatrix} -2 & -2 \\ 1 & 0 \end{vmatrix} &= (-2)(0) - (-2)(1) \\ &= 0 + 2 = 2 \\[4pt] \begin{vmatrix} -2 & -1 \\ 1 & -1 \end{vmatrix} &= (-2)(-1) - (-1)(1) \\ &= 2 + 1 = 3 \end{aligned}

Multiply by the entries and their signs, then add:

0(−2)=0−2(2)=−41(3)=30−4+3=−1\begin{aligned} 0(-2) &= 0 \\ -2(2) &= -4 \\ 1(3) &= 3 \\[4pt] 0 - 4 + 3 &= -1 \end{aligned}

So A−1A^{-1} exists. The cofactors of A are

Why?

A square matrix has an inverse exactly when its determinant is not zero. Write this reason down: examiners look for it.

A11=−2, A12=−2, A13=3A21=−1, A22=−1, A23=2A31=−3, A32=−2, A33=4\begin{aligned} A_{11} &= -2,\ A_{12} = -2,\ A_{13} = 3 \\ A_{21} &= -1,\ A_{22} = -1,\ A_{23} = 2 \\ A_{31} &= -3,\ A_{32} = -2,\ A_{33} = 4 \end{aligned}
Why?

Each cofactor is its minor with a sign:

Aij=(−1)i+jMijA_{ij} = (-1)^{i + j}M_{ij}

For example, deleting row 2 and column 1 leaves

M21=∣21−10∣=1M_{21} = \begin{vmatrix} 2 & 1 \\ -1 & 0 \end{vmatrix} = 1

and 2+12 + 1 is odd, so A21=−1A_{21} = -1⁠.

The signs (−1)i+j(-1)^{i + j} follow the pattern

[+−+−+−+−+]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

on the entries of

A=[021−2−1−21−10]A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix}

Row 1 (cover row 1 and one column each time):

A11=+∣−1−2−10∣=+((−1)(0)−(−2)(−1))=+(0−2)=−2A12=−∣−2−210∣=−((−2)(0)−(−2)(1))=−(0+2)=−2A13=+∣−2−11−1∣=+((−2)(−1)−(−1)(1))=+(2+1)=3\begin{aligned} A_{11} &= +\begin{vmatrix} -1 & -2 \\ -1 & 0 \end{vmatrix} \\ &= +\big((-1)(0) - (-2)(-1)\big) \\ &= +(0 - 2) = -2 \\[4pt] A_{12} &= -\begin{vmatrix} -2 & -2 \\ 1 & 0 \end{vmatrix} \\ &= -\big((-2)(0) - (-2)(1)\big) \\ &= -(0 + 2) = -2 \\[4pt] A_{13} &= +\begin{vmatrix} -2 & -1 \\ 1 & -1 \end{vmatrix} \\ &= +\big((-2)(-1) - (-1)(1)\big) \\ &= +(2 + 1) = 3 \end{aligned}

Row 2 (cover row 2):

A21=−∣21−10∣=−((2)(0)−(1)(−1))=−(0+1)=−1A22=+∣0110∣=+((0)(0)−(1)(1))=+(0−1)=−1A23=−∣021−1∣=−((0)(−1)−(2)(1))=−(0−2)=−(−2)=2\begin{aligned} A_{21} &= -\begin{vmatrix} 2 & 1 \\ -1 & 0 \end{vmatrix} \\ &= -\big((2)(0) - (1)(-1)\big) \\ &= -(0 + 1) = -1 \\[4pt] A_{22} &= +\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \\ &= +\big((0)(0) - (1)(1)\big) \\ &= +(0 - 1) = -1 \\[4pt] A_{23} &= -\begin{vmatrix} 0 & 2 \\ 1 & -1 \end{vmatrix} \\ &= -\big((0)(-1) - (2)(1)\big) \\ &= -(0 - 2) = -(-2) = 2 \end{aligned}

Row 3 (cover row 3):

A31=+∣21−1−2∣=+((2)(−2)−(1)(−1))=+(−4+1)=−3A32=−∣01−2−2∣=−((0)(−2)−(1)(−2))=−(0+2)=−2A33=+∣02−2−1∣=+((0)(−1)−(2)(−2))=+(0+4)=4\begin{aligned} A_{31} &= +\begin{vmatrix} 2 & 1 \\ -1 & -2 \end{vmatrix} \\ &= +\big((2)(-2) - (1)(-1)\big) \\ &= +(-4 + 1) = -3 \\[4pt] A_{32} &= -\begin{vmatrix} 0 & 1 \\ -2 & -2 \end{vmatrix} \\ &= -\big((0)(-2) - (1)(-2)\big) \\ &= -(0 + 2) = -2 \\[4pt] A_{33} &= +\begin{vmatrix} 0 & 2 \\ -2 & -1 \end{vmatrix} \\ &= +\big((0)(-1) - (2)(-2)\big) \\ &= +(0 + 4) = 4 \end{aligned}

Watch out: the sign pattern starts with ++ at the top left and alternates.

adj⁡A=[−2−1−3−2−1−2324]\operatorname{adj}A = \begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix}
Why?

adj A is the transpose of the matrix of cofactors: the cofactors of row 1 of A form column 1 of adj A, and so on.

Column 1: A11,A12,A13=−2,−2,3Column 2: A21,A22,A23=−1,−1,2Column 3: A31,A32,A33=−3,−2,4\begin{aligned} &\text{Column 1: } A_{11}, A_{12}, A_{13} = -2, -2, 3 \\ &\text{Column 2: } A_{21}, A_{22}, A_{23} = -1, -1, 2 \\ &\text{Column 3: } A_{31}, A_{32}, A_{33} = -3, -2, 4 \end{aligned}
A−1=1∣A∣adj⁡A=[213212−3−2−4]\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \begin{bmatrix} 2 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix} \end{aligned}
Why?

Here 1∣A∣=−1\frac{1}{|A|} = -1⁠, so every entry of adj A changes sign.

A−1=1−1[−2−1−3−2−1−2324]=−1[−2−1−3−2−1−2324]\begin{aligned} A^{-1} &= \frac{1}{-1}\begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} \\ &= -1\begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} \end{aligned}

The given system can be written as ATX=BA^TX = B⁠, where

AT=[0−212−1−11−20],A^T = \begin{bmatrix} 0 & -2 & 1 \\ 2 & -1 & -1 \\ 1 & -2 & 0 \end{bmatrix}, X=[xyz], B=[7810]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix}
Why?

Write the coefficients of x,y,zx, y, z of each equation as a row. −2y+z=7-2y + z = 7 gives (0,−2,1)(0, -2, 1)⁠, 2x−y−z=82x - y - z = 8 gives (2,−1,−1)(2, -1, -1)⁠, and x−2y=10x - 2y = 10 gives (1,−2,0)(1, -2, 0)⁠.

These rows are the columns of A, so the coefficient matrix is ATA^T⁠, not A.

∣AT∣=∣A∣≠0|A^T| = |A| \ne 0⁠, so the system has a unique solution X=(AT)−1B=(A−1)TBX = (A^T)^{-1}B = (A^{-1})^TB⁠.

Why?

Two facts about the transpose:

∣AT∣=∣A∣(AT)−1=(A−1)T\begin{aligned} |A^T| &= |A| \\ (A^T)^{-1} &= (A^{-1})^T \end{aligned}

So we can use the A−1A^{-1} already found: just transpose it. Row 1 of A−1A^{-1}⁠, (2,1,3)(2, 1, 3)⁠, becomes column 1 of (A−1)T(A^{-1})^T⁠, and so on:

(A−1)T=[22−311−232−4](A^{-1})^T = \begin{bmatrix} 2 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix}
[xyz]=[22−311−232−4][7810]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix}\begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} =[14+16−307+8−2021+16−40]=[0−5−3]= \begin{bmatrix} 14 + 16 - 30 \\ 7 + 8 - 20 \\ 21 + 16 - 40 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix}
Why?

Each entry is a row of (A−1)T(A^{-1})^T times the column B. The first is 2(7)+2(8)−3(10)2(7) + 2(8) - 3(10)⁠.

Row 1: (2)(7)+(2)(8)+(−3)(10)=14+16−30=30−30=0Row 2: (1)(7)+(1)(8)+(−2)(10)=7+8−20=15−20=−5Row 3: (3)(7)+(2)(8)+(−4)(10)=21+16−40=37−40=−3\begin{aligned} &\text{Row 1: } (2)(7) + (2)(8) + (-3)(10) \\ &= 14 + 16 - 30 \\ &= 30 - 30 = 0 \\[4pt] &\text{Row 2: } (1)(7) + (1)(8) + (-2)(10) \\ &= 7 + 8 - 20 \\ &= 15 - 20 = -5 \\[4pt] &\text{Row 3: } (3)(7) + (2)(8) + (-4)(10) \\ &= 21 + 16 - 40 \\ &= 37 - 40 = -3 \end{aligned}

Check in the given equations: −2(−5)+(−3)=7-2(-5) + (-3) = 7⁠, 0+5+3=80 + 5 + 3 = 8 and 0+10=100 + 10 = 10⁠.

−2y+z=−2(−5)+(−3)=10−3=72x−y−z=2(0)−(−5)−(−3)=0+5+3=8x−2y=0−2(−5)=0+10=10\begin{aligned} -2y + z &= -2(-5) + (-3) \\ &= 10 - 3 = 7 \\[4pt] 2x - y - z &= 2(0) - (-5) - (-3) \\ &= 0 + 5 + 3 = 8 \\[4pt] x - 2y &= 0 - 2(-5) \\ &= 0 + 10 = 10 \end{aligned}

∴ x=0\therefore\ x = 0⁠, y=−5y = -5⁠, z=−3z = -3⁠.

(b)

The matrix is singular, so its determinant is 0:

Why?

A singular matrix is one whose determinant is zero. This gives an equation in xx⁠.

∣3−1sin⁡3x−74cos⁡2x−1172∣=0\begin{vmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{vmatrix} = 0

Expanding along R1R_1⁠,

3(8−7cos⁡2x)+1(−14+11cos⁡2x)+sin⁡3x (−49+44)=0⇒ 10−10cos⁡2x−5sin⁡3x=0⇒ 2cos⁡2x+sin⁡3x=2\begin{aligned} &3(8 - 7\cos 2x) \\ &\quad + 1(-14 + 11\cos 2x) \\ &\quad + \sin 3x\,(-49 + 44) = 0 \\ \Rightarrow\ &10 - 10\cos 2x - 5\sin 3x = 0 \\ \Rightarrow\ &2\cos 2x + \sin 3x = 2 \end{aligned}
Why?

The three minors along R1R_1 are

4(2)−7cos⁡2x(−7)(2)−(−11)cos⁡2x(−7)(7)−4(−11)\begin{aligned} &4(2) - 7\cos 2x \\ &(-7)(2) - (-11)\cos 2x \\ &(-7)(7) - 4(-11) \end{aligned}

Each one, worked out:

∣4cos⁡2x72∣=(4)(2)−(cos⁡2x)(7)=8−7cos⁡2x∣−7cos⁡2x−112∣=(−7)(2)−(cos⁡2x)(−11)=−14+11cos⁡2x∣−74−117∣=(−7)(7)−(4)(−11)=−49+44=−5\begin{aligned} &\begin{vmatrix} 4 & \cos 2x \\ 7 & 2 \end{vmatrix} \\ &= (4)(2) - (\cos 2x)(7) \\ &= 8 - 7\cos 2x \\[4pt] &\begin{vmatrix} -7 & \cos 2x \\ -11 & 2 \end{vmatrix} \\ &= (-7)(2) - (\cos 2x)(-11) \\ &= -14 + 11\cos 2x \\[4pt] &\begin{vmatrix} -7 & 4 \\ -11 & 7 \end{vmatrix} \\ &= (-7)(7) - (4)(-11) \\ &= -49 + 44 = -5 \end{aligned}

The middle entry is −1-1 and its sign is −-⁠, so it enters as +1+1⁠.

Multiply each minor by its entry and sign:

3(8−7cos⁡2x)=24−21cos⁡2x1(−14+11cos⁡2x)=−14+11cos⁡2xsin⁡3x (−5)=−5sin⁡3x\begin{aligned} &3(8 - 7\cos 2x) \\ &= 24 - 21\cos 2x \\[4pt] &1(-14 + 11\cos 2x) \\ &= -14 + 11\cos 2x \\[4pt] &\sin 3x\,(-5) = -5\sin 3x \end{aligned}

Collect: 24−14=1024 - 14 = 10⁠, and the cos⁡2x\cos 2x terms give −21+11=−10-21 + 11 = -10⁠. Then divide by −5-5⁠.

Dividing each term by −5-5⁠: 10−5=−2\frac{10}{-5} = -2⁠, −10cos⁡2x−5=2cos⁡2x\frac{-10\cos 2x}{-5} = 2\cos 2x⁠, −5sin⁡3x−5=sin⁡3x\frac{-5\sin 3x}{-5} = \sin 3x⁠.

10−10cos⁡2x−5sin⁡3x=0−2+2cos⁡2x+sin⁡3x=02cos⁡2x+sin⁡3x=2\begin{aligned} &10 - 10\cos 2x - 5\sin 3x = 0 \\ &-2 + 2\cos 2x + \sin 3x = 0 \\ &2\cos 2x + \sin 3x = 2 \end{aligned}

Using cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2x and sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x⁠,

Why?

The equation mixes cos⁡2x\cos 2x and sin⁡3x\sin 3x⁠. Both can be written in terms of sin⁡x\sin x alone, which turns the equation into a polynomial in sin⁡x\sin x⁠.

2−4sin⁡2x+3sin⁡x−4sin⁡3x=2⇒ 4sin⁡3x+4sin⁡2x−3sin⁡x=0⇒ sin⁡x(4sin⁡2x+4sin⁡x−3)=0\begin{aligned} &2 - 4\sin^2x + 3\sin x \\ &\quad - 4\sin^3x = 2 \\ \Rightarrow\ &4\sin^3x + 4\sin^2x - 3\sin x = 0 \\ \Rightarrow\ &\sin x(4\sin^2x + 4\sin x - 3) = 0 \end{aligned}
Why?

2cos⁡2x=2−4sin⁡2x2\cos 2x = 2 - 4\sin^2x⁠. The 2 cancels with the 2 on the right. Multiply by −1-1⁠, then take sin⁡x\sin x out as a common factor.

2cos⁡2x=2(1−2sin⁡2x)=2−4sin⁡2x2−4sin⁡2x+3sin⁡x−4sin⁡3x=2−4sin⁡2x+3sin⁡x−4sin⁡3x=04sin⁡3x+4sin⁡2x−3sin⁡x=0sin⁡x(4sin⁡2x+4sin⁡x−3)=0\begin{aligned} &2\cos 2x = 2(1 - 2\sin^2x) \\ &= 2 - 4\sin^2x \\[4pt] &2 - 4\sin^2x + 3\sin x \\ &\quad - 4\sin^3x = 2 \\[4pt] &-4\sin^2x + 3\sin x \\ &\quad - 4\sin^3x = 0 \\[4pt] &4\sin^3x + 4\sin^2x \\ &\quad - 3\sin x = 0 \\[4pt] &\sin x(4\sin^2x + 4\sin x - 3) = 0 \end{aligned}

Watch out: do not divide by sin⁡x\sin x⁠. That would lose the solutions with sin⁡x=0\sin x = 0⁠.

Factorising,

sin⁡x(2sin⁡x−1)(2sin⁡x+3)=0\sin x(2\sin x - 1)(2\sin x + 3) = 0
Why?

Split the middle term: we need two numbers with product 4(−3)=−124(-3) = -12 and sum 4, which are 6 and −2-2⁠.

4s2+6s−2s−3=2s(2s+3)−(2s+3)=(2s−1)(2s+3)\begin{aligned} &4s^2 + 6s - 2s - 3 \\ &= 2s(2s + 3) - (2s + 3) \\ &= (2s - 1)(2s + 3) \end{aligned}

with s=sin⁡xs = \sin x⁠.

sin⁡x=−32\sin x = -\frac32 is not possible. So sin⁡x=0\sin x = 0 or sin⁡x=12\sin x = \frac12⁠.

Why?

sin⁡x\sin x always lies between −1-1 and 1, and −32<−1-\frac32 < -1⁠.

Set each factor equal to 0:

sin⁡x=02sin⁡x−1=02sin⁡x=1sin⁡x=122sin⁡x+3=02sin⁡x=−3sin⁡x=−32\begin{aligned} \sin x &= 0 \\[4pt] 2\sin x - 1 &= 0 \\ 2\sin x &= 1 \\ \sin x &= \tfrac12 \\[4pt] 2\sin x + 3 &= 0 \\ 2\sin x &= -3 \\ \sin x &= -\tfrac32 \end{aligned}

In [0,π2]\left[0, \frac{\pi}{2}\right]⁠, sin⁡0=0\sin 0 = 0 and sin⁡π6=12\sin\frac{\pi}{6} = \frac12⁠.

For x∈[0,π2]x \in \left[0, \frac{\pi}{2}\right]⁠: sin⁡x=0⇒x=0\sin x = 0 \Rightarrow x = 0⁠, and sin⁡x=12⇒x=π6\sin x = \frac12 \Rightarrow x = \frac{\pi}{6}⁠.

∴ x=0\therefore\ x = 0 or x=π6x = \dfrac{\pi}{6}⁠.

Where this comes from

Chapter 4, Determinants: Inverse by adjoint, linear system; singular matrix.

Question type: (a) Solve equations by the matrix method; (b) Evaluate a determinant

How to spot it: click to see

(a) Type 9. Solve equations by the matrix method

Spot it: "solve, using the matrix method", with two or three equations. A word problem (prices, three numbers) is the same: write the equations first.

Here: Find A−1A^{-1} first: ∣A∣=−1≠0|A| = -1 \ne 0⁠, so A−1=−adj⁡AA^{-1} = -\operatorname{adj}A⁠. The coefficient rows of the equations are the columns of AA⁠, so the system is ATX=BA^TX = B and X=(A−1)TBX = (A^{-1})^TB⁠.

(b) Type 1. Evaluate a determinant: cross for 2×22\times2⁠, expand for 3×33\times3

Spot it: "evaluate" and a square block of numbers between straight bars.

Here: Singular means the determinant is 0: expand the 3×33 \times 3 along R1R_1 to get 10−10cos⁡2x−5sin⁡3x=010 - 10\cos 2x - 5\sin 3x = 0⁠, then write it in sin⁡x\sin x and factorise.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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